Obfuscated y=x, I took the graph of the normal curve, shifted it right, reduced the standard deviation, took the derivative of the curve and found the equation for the line tangent to the curve at the point below. I think it works. https://www.desmos.com/calculator/ldp8mnypnt
First!
…
y = x
y=(x^3+x)/(x^2+1)
It looks like y=x
And it is! Can you come up with a different function that has the same graph?
https://www.desmos.com/calculator/hn9zabgjwt
Ok, here’s my second answer!
parametric equation: x(t) = t, y(t) = t
https://www.desmos.com/calculator/pwzvvt0krb
y=2(1/2(x)-1/2) + 1. Not very creative, but definitely not the same as y=x.
x(t)=tan(2t)
y(t)=2tan(t)/(1-tan^2(t))
-pi/4<t<pi/4 (one period) but since Desmos doesn't allow pi for t I chose -.8<=t<=.8 (covers slightly more than one full period).
https://www.desmos.com/calculator/3f84ozgdvx
Here’s another! https://www.desmos.com/calculator/zqtypvrcja
y = tan(arctan(x))
Would that create holes at x = (2n+1)(pi/2), for n integer?
Here’s one: https://www.desmos.com/calculator/1xtzaono7e
y = floor(x) + mod(x,1)
Mine is a bit /mean/ since I did leave off a few people’s responses. https://www.desmos.com/calculator/x5qs01vz9a
Very creative. Here’s my not so creative answer: https://www.desmos.com/calculator/v5d2n2b2jj
Gauntlet thrown down! calculus, statistics, summation, PI notation, taylor series, etc.
https://www.desmos.com/calculator/r5az8lbyzm
Theta = pi/4 in polar form
Got another here: https://www.desmos.com/calculator/azwovxy77y
Nothing like a little differentiation and trig substitution!
Obfuscated y=x, I took the graph of the normal curve, shifted it right, reduced the standard deviation, took the derivative of the curve and found the equation for the line tangent to the curve at the point below. I think it works.
https://www.desmos.com/calculator/ldp8mnypnt
No doubt it is y=x …perfect!
y=x^(x^0)
Nicely done Laila!
Would that yield a removable discontinuity at (0,0)?
How about y=d(.5x^2+6)/dx ?
Great! First calculus answer??